3.224 \(\int (d \cos (a+b x))^{5/2} \csc (a+b x) \, dx\)

Optimal. Leaf size=78 \[ \frac {d^{5/2} \tan ^{-1}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b}-\frac {d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b}+\frac {2 d (d \cos (a+b x))^{3/2}}{3 b} \]

[Out]

d^(5/2)*arctan((d*cos(b*x+a))^(1/2)/d^(1/2))/b-d^(5/2)*arctanh((d*cos(b*x+a))^(1/2)/d^(1/2))/b+2/3*d*(d*cos(b*
x+a))^(3/2)/b

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Rubi [A]  time = 0.06, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {2565, 321, 329, 298, 203, 206} \[ \frac {d^{5/2} \tan ^{-1}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b}-\frac {d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b}+\frac {2 d (d \cos (a+b x))^{3/2}}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[(d*Cos[a + b*x])^(5/2)*Csc[a + b*x],x]

[Out]

(d^(5/2)*ArcTan[Sqrt[d*Cos[a + b*x]]/Sqrt[d]])/b - (d^(5/2)*ArcTanh[Sqrt[d*Cos[a + b*x]]/Sqrt[d]])/b + (2*d*(d
*Cos[a + b*x])^(3/2))/(3*b)

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rubi steps

\begin {align*} \int (d \cos (a+b x))^{5/2} \csc (a+b x) \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {x^{5/2}}{1-\frac {x^2}{d^2}} \, dx,x,d \cos (a+b x)\right )}{b d}\\ &=\frac {2 d (d \cos (a+b x))^{3/2}}{3 b}-\frac {d \operatorname {Subst}\left (\int \frac {\sqrt {x}}{1-\frac {x^2}{d^2}} \, dx,x,d \cos (a+b x)\right )}{b}\\ &=\frac {2 d (d \cos (a+b x))^{3/2}}{3 b}-\frac {(2 d) \operatorname {Subst}\left (\int \frac {x^2}{1-\frac {x^4}{d^2}} \, dx,x,\sqrt {d \cos (a+b x)}\right )}{b}\\ &=\frac {2 d (d \cos (a+b x))^{3/2}}{3 b}-\frac {d^3 \operatorname {Subst}\left (\int \frac {1}{d-x^2} \, dx,x,\sqrt {d \cos (a+b x)}\right )}{b}+\frac {d^3 \operatorname {Subst}\left (\int \frac {1}{d+x^2} \, dx,x,\sqrt {d \cos (a+b x)}\right )}{b}\\ &=\frac {d^{5/2} \tan ^{-1}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b}-\frac {d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b}+\frac {2 d (d \cos (a+b x))^{3/2}}{3 b}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 68, normalized size = 0.87 \[ \frac {(d \cos (a+b x))^{5/2} \left (2 \cos ^{\frac {3}{2}}(a+b x)+3 \tan ^{-1}\left (\sqrt {\cos (a+b x)}\right )-3 \tanh ^{-1}\left (\sqrt {\cos (a+b x)}\right )\right )}{3 b \cos ^{\frac {5}{2}}(a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Cos[a + b*x])^(5/2)*Csc[a + b*x],x]

[Out]

((d*Cos[a + b*x])^(5/2)*(3*ArcTan[Sqrt[Cos[a + b*x]]] - 3*ArcTanh[Sqrt[Cos[a + b*x]]] + 2*Cos[a + b*x]^(3/2)))
/(3*b*Cos[a + b*x]^(5/2))

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fricas [B]  time = 0.60, size = 281, normalized size = 3.60 \[ \left [\frac {6 \, \sqrt {-d} d^{2} \arctan \left (\frac {2 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {-d}}{d \cos \left (b x + a\right ) + d}\right ) + 8 \, \sqrt {d \cos \left (b x + a\right )} d^{2} \cos \left (b x + a\right ) + 3 \, \sqrt {-d} d^{2} \log \left (-\frac {d \cos \left (b x + a\right )^{2} + 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {-d} {\left (\cos \left (b x + a\right ) - 1\right )} - 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1}\right )}{12 \, b}, -\frac {6 \, d^{\frac {5}{2}} \arctan \left (\frac {2 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {d}}{d \cos \left (b x + a\right ) - d}\right ) - 8 \, \sqrt {d \cos \left (b x + a\right )} d^{2} \cos \left (b x + a\right ) - 3 \, d^{\frac {5}{2}} \log \left (-\frac {d \cos \left (b x + a\right )^{2} - 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {d} {\left (\cos \left (b x + a\right ) + 1\right )} + 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1}\right )}{12 \, b}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(b*x+a))^(5/2)*csc(b*x+a),x, algorithm="fricas")

[Out]

[1/12*(6*sqrt(-d)*d^2*arctan(2*sqrt(d*cos(b*x + a))*sqrt(-d)/(d*cos(b*x + a) + d)) + 8*sqrt(d*cos(b*x + a))*d^
2*cos(b*x + a) + 3*sqrt(-d)*d^2*log(-(d*cos(b*x + a)^2 + 4*sqrt(d*cos(b*x + a))*sqrt(-d)*(cos(b*x + a) - 1) -
6*d*cos(b*x + a) + d)/(cos(b*x + a)^2 + 2*cos(b*x + a) + 1)))/b, -1/12*(6*d^(5/2)*arctan(2*sqrt(d*cos(b*x + a)
)*sqrt(d)/(d*cos(b*x + a) - d)) - 8*sqrt(d*cos(b*x + a))*d^2*cos(b*x + a) - 3*d^(5/2)*log(-(d*cos(b*x + a)^2 -
4*sqrt(d*cos(b*x + a))*sqrt(d)*(cos(b*x + a) + 1) + 6*d*cos(b*x + a) + d)/(cos(b*x + a)^2 - 2*cos(b*x + a) +
1)))/b]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}} \csc \left (b x + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(b*x+a))^(5/2)*csc(b*x+a),x, algorithm="giac")

[Out]

integrate((d*cos(b*x + a))^(5/2)*csc(b*x + a), x)

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maple [B]  time = 0.18, size = 244, normalized size = 3.13 \[ -\frac {d^{\frac {5}{2}} \ln \left (\frac {2 \sqrt {d}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}-4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )+1}\right )}{2 b}-\frac {d^{\frac {5}{2}} \ln \left (\frac {2 \sqrt {d}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}+4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )-1}\right )}{2 b}-\frac {4 \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}\, d^{2} \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{3 b}+\frac {2 d^{2} \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}}{3 b}-\frac {d^{3} \ln \left (\frac {2 \sqrt {-d}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )}\right )}{\sqrt {-d}\, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*cos(b*x+a))^(5/2)*csc(b*x+a),x)

[Out]

-1/2/b*d^(5/2)*ln(2/(cos(1/2*b*x+1/2*a)+1)*(d^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)-2*d*cos(1/2*b*x+1/2*a)
-d))-1/2/b*d^(5/2)*ln(2/(cos(1/2*b*x+1/2*a)-1)*(d^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)+2*d*cos(1/2*b*x+1/
2*a)-d))-4/3/b*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)*d^2*sin(1/2*b*x+1/2*a)^2+2/3/b*d^2*(-2*sin(1/2*b*x+1/2*a)^2
*d+d)^(1/2)-1/(-d)^(1/2)/b*d^3*ln(2/cos(1/2*b*x+1/2*a)*((-d)^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)-d))

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maxima [A]  time = 0.60, size = 83, normalized size = 1.06 \[ \frac {6 \, d^{\frac {7}{2}} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )}}{\sqrt {d}}\right ) + 3 \, d^{\frac {7}{2}} \log \left (\frac {\sqrt {d \cos \left (b x + a\right )} - \sqrt {d}}{\sqrt {d \cos \left (b x + a\right )} + \sqrt {d}}\right ) + 4 \, \left (d \cos \left (b x + a\right )\right )^{\frac {3}{2}} d^{2}}{6 \, b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(b*x+a))^(5/2)*csc(b*x+a),x, algorithm="maxima")

[Out]

1/6*(6*d^(7/2)*arctan(sqrt(d*cos(b*x + a))/sqrt(d)) + 3*d^(7/2)*log((sqrt(d*cos(b*x + a)) - sqrt(d))/(sqrt(d*c
os(b*x + a)) + sqrt(d))) + 4*(d*cos(b*x + a))^(3/2)*d^2)/(b*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (d\,\cos \left (a+b\,x\right )\right )}^{5/2}}{\sin \left (a+b\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*cos(a + b*x))^(5/2)/sin(a + b*x),x)

[Out]

int((d*cos(a + b*x))^(5/2)/sin(a + b*x), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(b*x+a))**(5/2)*csc(b*x+a),x)

[Out]

Timed out

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